Ethyl butanoate, CH3CH₂CH₂CO₂CH₂CH3, is one of the many organic compounds isolated from mangoes. Which hydrogen is most readily removed when ethyl butanoate is treated with base? Propose a reason for your choice, and using the data listed below, estimate its pK₁. Compound CH3CONH2 CH3CHO pKa 16 17

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i stumbled my way to the right answersa in this question, but wanted to reinforce my understanding. could you please explain the step to solving these, especially the estimated pka at the end

Be sure to answer all parts.
Ethyl butanoate, CH3CH₂CH₂CO₂CH₂CH3, is one of the many organic compounds isolated from
mangoes. Which hydrogen is most readily removed when ethyl butanoate is treated with base?
Propose a reason for your choice, and using the data listed below, estimate its pK₁.
Compound
pka
CH3CONH2
16
CH3CHO
17
18
19.2
24.5
25
25
25
30
(CH3)3COH
(CH3)2C=O
CH3CO₂CH₂CH3
HC=CH
CH3C=N
CHCI 3
CH3CON(CH3)2
The most readily removed type of hydrogen is on the:
CH3 of the propyl group.
CH₂ bonded to C=O.
CH₂ of the ethyl group.
CH3 of the ethyl group.
This type of hydrogen is most readily removed because:
The estimated pK₂ is 24.5
Loss of the proton deceases steric hindrance around the C=O.
The conjugate base is stabilized by inductance.
The conjugate base is stabilized by resonance.
The conjugate base is stabilized by the high electronegativity of O.
Transcribed Image Text:Be sure to answer all parts. Ethyl butanoate, CH3CH₂CH₂CO₂CH₂CH3, is one of the many organic compounds isolated from mangoes. Which hydrogen is most readily removed when ethyl butanoate is treated with base? Propose a reason for your choice, and using the data listed below, estimate its pK₁. Compound pka CH3CONH2 16 CH3CHO 17 18 19.2 24.5 25 25 25 30 (CH3)3COH (CH3)2C=O CH3CO₂CH₂CH3 HC=CH CH3C=N CHCI 3 CH3CON(CH3)2 The most readily removed type of hydrogen is on the: CH3 of the propyl group. CH₂ bonded to C=O. CH₂ of the ethyl group. CH3 of the ethyl group. This type of hydrogen is most readily removed because: The estimated pK₂ is 24.5 Loss of the proton deceases steric hindrance around the C=O. The conjugate base is stabilized by inductance. The conjugate base is stabilized by resonance. The conjugate base is stabilized by the high electronegativity of O.
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